I'm thinking you want something that looks like this
[tex]\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}[/tex] or it could be like this
[tex]\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)[/tex]
use chain rule
dy/dx f(g(x))=f'(g(x))g'(x)
and √x=x^(1/2)
so
I'll do the first one first
[tex]\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}[/tex]=
[tex]\frac{dy}{dx}(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}[/tex]=
[tex]\frac{1}{2(x+\frac{1}{4}sin^2(2x))^{\frac{1}{2}}}(1+sin(2x)cos(2x))[/tex]=
[tex]\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}[/tex]
the 2nd one is a bit simpler
[tex]\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)[/tex]=
[tex]\frac{dy}{dx}x^{\frac{1}{2}}+\frac{1}{4}sin^2(2x)[/tex]=
[tex]\frac{1}{2}x^{\frac{-1}{2}}+\frac{1}{4}sin^2(2x)[/tex]=
[tex]\frac{1}{2x^{\frac{1}{2}}}+sin(2x)cos(2x)[/tex]=
[tex]\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)[/tex]
in conclusion
[tex]\frac{dy}{dx}\sqrt{x+\frac{1}{4}sin^2(2x)}[/tex]=[tex]\frac{1+sin(2x)cos(2x)}{2\sqrt{x+\frac{1}{4}sin^2(2x}}[/tex] and
[tex]\frac{dy}{dx}\sqrt{x}+\frac{1}{4}sin^2(2x)[/tex]=[tex]\frac{1}{2\sqrt{x}}+sin(2x)cos(2x)[/tex]
depends which one it was
